A taxi without any passenger moving on a frictionless horizontal Road with a velocity U can be stopped in a distance d. Now the passenger adds 40% to its weight. What is the stopping distance at velocity u, if the retardation remains the same?

Hint: in this question, as the taxi has initial velocity = U, and this taxi comes to rest so final velocity is V = 0. Then use Newton's equation i.e. ${v^2} - {u^2} = 2ad$, from this we find the distance d. now the weight increases so force will also increase i.e. $F = ma + \dfrac{{40ma}}{{100}}$ from this we find the acceleration a. After that we again use Newton’s equation to get the required stopping distance.

Complete step-by-step answer:
Let us consider a taxi without any passenger moving on a frictionless horizontal road with velocity u and can be stopped in a distance d.
Therefore, initial velocity = u
Final velocity v = 0
Using Newton’s equation of motion, we get
${v^2} - {u^2} = 2ad$
$ \Rightarrow {u^2} = 2ad$
$ \Rightarrow d = \dfrac{{{u^2}}}{{2a}}$ ……………………….. (1)
Let the retarding force is F
By Newton’s 2nd law of motion $F = ma$ …………………. (2)
Now, the passenger adds 40% to its weight. So, its force will also increase with its new acceleration ai, that is
$F = ma + {\dfrac{{40ma}}{{100}}^\prime }$
From equation (2), we get
$ \Rightarrow ma = ma + \dfrac{{40}}{{100}}ma'$
On solving we get the new acceleration
$ \Rightarrow a' = 1.4a$
Now again using the Newton’s equation of motion, we get
${v^2} - {u^2} = 2a's$
S is the stopping distance
Again, initial velocity u = 0 and final velocity = v, we get
$ \Rightarrow {u^2} = 2a's$
$ \Rightarrow s = \dfrac{{{u^2}}}{{2a'}}$
$ \Rightarrow s = \dfrac{{{u^2}}}{{2 \times 1.4a}} = \dfrac{1}{{1.4a}} \times \dfrac{{{u^2}}}{{2a}}$
Using equation (1), we get

$ \Rightarrow s = \dfrac{1}{{1.4}}d$
This is the required stopping distance.

Note: Retarding force is the resultant force, the force that is against the direction of the object's current velocity is the retarding force. The formula for retarding force, Retarding force is always negative as the direction of force and the direction of motion is opposite to each other, making 180° between them.